Specialization Math Example 2

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Example 2

medium

The AM-GM inequality states: for positive reals

a,b

,

\frac{a+b}{2} \ge \sqrt{ab}

. Specialise to

a = x^2

and

b = \frac{1}{x^2}

(for

x \ne 0

) and state what you get.

Solution

1
Substitute $a=x^2$ , $b=\frac{1}{x^2}$ (both positive for $x \ne 0$ ).
2
AM-GM gives: $\frac{x^2 + \frac{1}{x^2}}{2} \ge \sqrt{x^2 \cdot \frac{1}{x^2}} = \sqrt{1} = 1$ .
3
So $x^2 + \frac{1}{x^2} \ge 2$ for all $x \ne 0$ , with equality when $x^2 = \frac{1}{x^2}$ , i.e., $x = \pm 1$ .

Answer

x^2 + \frac{1}{x^2} \ge 2 \text{ for all } x \ne 0

Specialising the AM-GM inequality to chosen values of

a

and

b

produces many useful inequalities. The choice

a=x^2, b=1/x^2

is motivated by wanting a bound involving

x^2+1/x^2

.

About Specialization

Applying a general theorem or formula to a specific case by substituting particular values for the variables or parameters.

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Example 4 medium

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